Question

How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?

Answer 1

Hint: It is given in the question that we have to only look at 3 digit numbers because between 100 and 1000 there are only 3 digit numbers. Now we will use the fact that numbers should have only 2 and 9 as their digits, so we will find all the possible arrangements that can be done using these conditions and that will be the final answer.

Complete step-by-step answer:
Let’s start the solution,
Case 1: When we have three 2’s
Then the only possible number is 222.
Hence, one number in this case.
Case 2: When we have two 2’s and one 9
Now we will use the formula for arrangement of number like ‘aab’: $\dfrac{3!}{2!}=\dfrac{3\times 2}{2}=3$
The number of possible ways in which we can arrange this is: $\dfrac{3!}{2!}=\dfrac{3\times 2}{2}=3$
Hence, 3 numbers in this case.
Case 3: When we have one 2 and two 9’s
Now we will use the formula for arrangement of number like ‘aab’: $\dfrac{3!}{2!}=\dfrac{3\times 2}{2}=3$
The number of possible ways in which we can arrange this is: $\dfrac{3!}{2!}=\dfrac{3\times 2}{2}=3$
Hence, 3 numbers in this case.
Case 4: When we have three 9’s
Then the only possible number is 999.
Hence, one number in this case.
Hence, the total numbers will be 1 + 3 + 3 + 1 = 8.

Note: From the language of the question one should be able to deduce that we only have to look at 3 digits numbers only. And one can also try to write all the possible numbers that only contain 2 and 9, and then we just have to add all the total possible numbers to get the answer.